Both solutions to a quadratic make sense -- looking for applications

Question

I'm looking for reasonably real, non-abstract applications modeled by quadratic equations where both solutions make sense. I'd like them to be accessible to high school algebra students.

One I come up with is firing a bullet through a sphere -- the two solutions correspond to the entry and exit points of the bullet, that is the intersections of the bullet's line and the sphere's surface. To keep it simple, we could do this instead with a line and circle in the xy plane, but the idea is the same.

Applications outside of physics would be particularly welcome.

Answer 1

Throw a ball upwards, you wanna know how long until it reaches some height?
Solve a quadratic, the two solutions correspond to when its on its way up and when its coming down again.

Answer 2

The simplest question I can think of is to characterize rectangles with side lengths $x, y$ and fixed perimeter $2x + 2y = p$ and area $xy = A$. The two roots of the corresponding quadratic equation (say in $x$), when they exist, correspond to the "tall" solution and the "wide" solution. But note that the statement of the problem is symmetric in $x$ and $y$, so if you know that a solution exists with $x \neq y$ then from one solution you automatically deduce the other. I think this is a nice example for students to see.

Answer 3

An application from number theory is elementary recurrence relations, with the Fibonacci numbers providing the easy example: the equation $F_{n+2} = F_{n+1}+F_n$ ties neatly into the quadratic equation $x^2-x-1=0$, and the Fibonacci numbers themselves turn out to be a linear combination of powers of the two solutions of this equation, $\phi$ and $-\phi^{-1}$. (The fact that the second solution is 'small' means that its contribution tends to $0$, but this isn't always the case; for instance, the equation $a_{n+2} = 5a_{n+1}-6a_{n}$ has as its 'core' solutions $2^n$ and $3^n$ and the general solution is a linear combination of these).

Answer 4

Knowing two sides $a$ and $c$ of a triangle and an angle $\alpha$ not between them—i.e. the SSA condition—does not determine the length of the third side $b$. See Figure 3 on this page. (Unfortunately, the convention at the length of the side opposite the vertex $A$ is denoted $a$ is going to make the notation a little confusing below.)

Set the origin at the vertex $A$ with the $x$-axis along $AC$. Then the vertex $B$ is at $(c \cos \alpha, c \sin \alpha)$ and $C$ is at $(b,0)$, and we have $$(c \cos \alpha - b)^2 + (c \sin \alpha - 0)^2 = a^{2},$$ which is a quadratic in $b$ both of whose solutions evidently make sense.